The two questions at hand are :
1.) Show that a $k$-regular graph of girth four has at least $2k$ vertices.
2.) Show that a $k$-regular graph of girth five has at least $k^{2} + 1$ vertices.
The standard solutions are given below:
Standard 1: Take one vertex and its k neighbors, together k + 1 vertices. These neighbors are not connected, since otherwise there would be a triangle. Thus, if we look at one of these neighbors, it has $k − 1$ neighbors that are not among the $k + 1$ we counted so far. Altogether we have $1 + k + (k − 1) = 2k$ vertices.
Standard 2: Let $G$ be a $k$-regular graph with girth five; thus there are no loops or parallel edges and $G$ is simple. Now for $u_{i},u_{j} \in N_{G}(v)$ (with $N_{G}(v) = \{u_{1},\ldots,u_{k}\}$) we have that $u_{i} \notin N_{G}(u_{j})$ and $u_{j} \notin N_{G}(u_{i})$ because otherwise we would have a cycle of length three on $v$. Additionally we have that $N_{G}(u_{i})/\{v\} \cap N_{G}(u_{j})/\{v\} = \{∅\}$ because otherwise there would be a cycle of length four on $v$ (note that $|V(N_{G}(u_{i})/\{v\})| = k-1$). Thus:
$$|V(G)| \geq 1 + |V(N_{G}(v))| + \sum_{i=1}^{k}|V(N_{G}(u_{i})/\{v\})| = 1 + k + (k-1)k = k^{2}+1$$
Now I came up with my own solution to the 1st problem, shown below:
Custom 1: Let $G$ be a $k$-regular graph of girth four; thus $G$ is simple. Let $P \subseteq G$ be the longest path in $G$; denote $P$ by $P = v_{1},e(v_{1},v_{2}),v_{2},\ldots, v_{n}$. Since $P$ is the longest path we have that $v \in N_{G}(v_{1}) \Rightarrow v \in V(P)$; denote the elements of $N_{G}(v_{1})$ by $v_{11},v_{12},\ldots, v_{1k}$, where the ordering corresponds to their order of appearance in $P$. We can denote $P$ by $P = v_{1},e(v_{1},v_{11}),v_{11},\ldots, v_{1i},\ldots, v_{1k},\ldots, v_{n}$. Since $G$ has girth four, we know that any $P_{i} \subseteq P$ with $P_{i} = v_{1i},e(v_{1i},v_{j}),v_{j},\ldots, v_{1(i+1)}$ is s.t. $|V(P_{i})| \geq 3$. There are $k$ neighbors, and since $P$ is a path we have (excluding $v_{1k}$ and $v_{1}$) that $|V(P)| \geq 2k-2$, and thus $V(G) \geq 2k$.
My question is: Why can't I use the same framework of my custom proof to say that $|V(P)| \geq 3k-1$ for graphs of girth five. I mean the standard proof makes perfect sense, but I don't see why my custom proof framework would only hold for graphs of girth 4; is it wrong?